Question

A 1kW carrier is modulated to a depth of 80%. The total power in the modulated wave is
A. 1.32
B. 1.566
C. 1.84
D. 1.96

Answer 1

we are given the power of a carrier wave. We are asked to find the power of the modulated when the carrier is modulated to a depth of 80%. We know the equation used to find the total power in a modulated wave. By substituting known values in the equation we will get the solution.

Formula used:
Power of modulated wave, ${{P}_{m}}=P\left( 1+\dfrac{{{\mu }^{2}}}{2} \right)$

Complete step-by-step answer:
In the question we are given the power of a carrier.
Power, P = 1 KW.
It is said that this carrier is modulated to a depth of 80%.
From this we can understand that the efficiency of the carrier is 80%.
Let us represent efficiency of the carrier by ‘$\mu$’, then we have the efficiency as,
$\mu =80$
Now we need to find the total power in the modulated wave.
We know that the equation to find total power in the modulated wave is given by,
${{P}_{m}}=P\left( 1+\dfrac{{{\mu }^{2}}}{2} \right)$, were ‘${{P}_{m}}$’ is the power of the modulated wave, ‘P’ is the power of the carrier and
‘$\mu$’ is efficient.
Since we know the values of power of carrier and efficiency of the carrier, we can substitute them in the above equation.
Thus we get,
$\Rightarrow {{P}_{m}}=1\left( 1+\dfrac{{{\left( \dfrac{80}{100} \right)}^{2}}}{2} \right)$
$\Rightarrow {{P}_{m}}=1\left( 1+\dfrac{0.64}{2} \right)$
$\Rightarrow {{P}_{m}}=1\times \left( 1+0.32 \right)$
$\Rightarrow {{P}_{m}}=1\times 1.32$
$\Rightarrow {{P}_{m}}=1.32KW$
Therefore we have the power of the modulated wave as 1.32 KW.

So, the correct answer is “Option A”.

Note: A carrier wave is a wave with fixed amplitude and frequency. We modulate the amplitude, frequency or phase of a carrier to carry signals.
When we modulate the amplitude it is called amplitude modulation (AM). When the frequency of the carrier wave is modified, it is frequency modulation (FM) and when we modify the phase of the wave, it is phase modulation (PM).